Computer Aided Analysis of Balancing of Multi-Cylinder Radial and V- Engines 1 2 D

ISSN 2319-8885 Vol.04,Issue.57, December-2015,

Pages:12325-12340

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Computer Aided Analysis of Balancing of Multi-Cylinder Radial and V- Engines 1 2 D. AMRUTHA VIJAY , SNEHALATHA. P 1PG Scholar, Eswar College of Engineering, Narasaraopet, Guntur, AP, India, E-mail: [email protected]. 2Assistant Professor, Eswar College of Engineering, Narasaraopet, Guntur, AP, India, E-mail: [email protected].

Abstract: The reciprocating engines are the prime source of power generation in various mechanical applications ranging from power generation to automobiles. Due to unbalanced moments and inertia forces cause noise, vibrations and harshness are caused in this engine which further cause complications in their operation. To minimize the unbalance, analysis for the unbalanced forces and moments for different configuration of cylinders and firing order of multi-cylinder radial and V engines are done. To minimize the time and calculations the C programs have been developed for this analysis. On the basis of resultant unbalanced forces and moments the radial, V engines and combined radial and inline configured engines are compared.

Keywords: Computer Aided Analysis, Multi-Cylinder Radial, V- Engines.

I. INTRODUCTION and aircrafts these varieties of engines are usually employed. A reciprocating engine may be a heat engine that uses one The V configurations engines are well-balanced and smooth. or over one reciprocating pistons to convert pressure into a Counterweights on the crankshaft are mainly used for the rotating motion. These engines are very widely used in balancing of the V10 and cross plane V8 engine. V12 engines extensively in cars, power generators and aircrafts. Multi- always have even firing and exceptional balance irrespective cylinder engines use over one cylinder that is employed of V angle. The radial engine could be told as a reciprocating extensively today. These engines will be a combination of any kind IC engine during which the cylinders purpose outward 2 kinds of engines, or inline, v-type, radial. In four- cylinder from a central crankshaft rather like the spokes on a wheel. and six-cylinder configurations inline engines are typically This configuration was widely employed in giant aircraft employed, with all cylinders aligned in one row, with no engines before the begin of turbine engines. In an exceedingly offset. In vehicles, locomotives and aircraft they're getting radial engine, master-and-articulating-rod assembly is used, though the word in-line incorporates a broader which employed to attach the pistons to the crankshaft. Four-stroke means in case of aircraft engines. An identical horizontally radial engines continuously have an odd variety of cylinders opposed or V engine is difficult to construct than a inline per row to urge an identical every-other-piston firing order engine, as a result of each the cylinder bank and crankshaft ensuing swish operation that is achieved by the engine taking are often milled from one metal casting, and fewer cylinder 2 revolutions of the crankshaft to complete the four strokes. heads and camshafts are needed. In-line engines are smaller in As a result there's continuously a two-piston gap between the overall physical dimensions over the styles just like the radial piston on its power stroke and therefore the next piston on its but have bigger benefits, and might be mounted in any compression stroke. On the opposite hand in case of a good direction. When compared to counterparts of V-shaped the variety of cylinders a three-piston gap between firing pistons straight configurations are easier. As compared to "flat and V" on the primary crank shaft revolution , and solely a one-piston engines that have support bearings between each 2 pistons gap on the second crank shaft revolution is observed, that Inline engines have a support bearing between every piston. A ends up in an uneven firing order at intervals the engine. V engine is seen as a general configuration in case of an These engines are well balanced. There’s no unbalanced enclosed combustion engine. The pistons and cylinders are torque within the plane of radial engines. separated by an angle known as v angle and aligned, in 2 separate planes or 'banks' so they seem to be in an Nowadays multi row radial engines have replaced the exceedingly "V" when viewed along the axis of the single row radial engine in case of large aero engine due to crankshaft. certain advantages. In multi row radial engines multiple no. Of radial engines with some offset distance are connected by Compared to a similar inline configuration the V a common shaft. Here we can get even number of cylinders configuration helps in reducing the general engine length, by the combination of radial engines with odd number of height and weight. V angle are employed in totally different cylinders arranged in even number of rows. These engines are engines in a numerous way; that depends on the amount of considered when high power is required. These engines are cylinders. 600, 450,300,900 are widely used V angles. In cars also well balanced compared to their counterpart single row

Copyright @ 2015 IJSETR. All rights reserved. D. AMRUTHA VIJAY, SNEHALATHA. P radial engines. The bigger advantage of this type of engines be in dynamic balance when there doesn’t exist any resultant over radial engines is: these types of engines can be easily centrifugal force in addition a resultant couple. cooled as compactness is less. 3. Balancing Of Reciprocating Masses: Acceleration of the II. BALANCING reciprocating mass of a slider-crank mechanism is given by The inertia forces related to moving mass present in rotary the equation Therefore, the forces required to accelerate mass or reciprocating machinery manufacture unbalance of force. m is, Balancing is that the method of planning or modifying machinery so the unbalance is reduced to a permissible level F= m rω2 [cosθ + (cos2θ)/n] and if doable is eliminated entirely. The time varying unbalance forces exerted on the frame by the moving machine = m rω2cosθ + m rω2 (cos2θ)/n provides vibrating motion to border and leads to noise production thus so as to avoid the catastrophic failure of machine caused owing to noise, vibration and harshness. The foremost common approach for balancing is by redistributing the lots which can be accomplished by adding or removing some quantity of mass from numerous machine members. 2 Maximum value of primary force = mrω Maximum value There are 2 basic styles of unbalance, reciprocating unbalance 2 of secondary force = mrω /n. and rotating unbalance, which can occur individually or together. Where n=L/R is much greater than unity; hence the secondary force is smaller compared to primary force and can be safely A. Balancing Of Rotating Masses 1. Static Balancing:A system of rotating masses is claimed to ignored for slow speed engines. be in static balance when the combined centre of mass of the 4. Primary And Secondary Balance: Primary balance is that system lies on the axis of rotation. the balance achieved by compensating for the eccentricities of the rotating masses, together with the connecting rods. Primary balance is controlled by addition or removal of mass to or from the crankshaft, at every end, at the required radius and angle that changes both owing to style and manufacturing tolerances. Theoretically any typical engine will be balanced utterly for primary balance.

Secondary balance is attained by compensating partially or fully for:  Kinetic energy of the pistons.  Non-sinusoidal motion of the pistons.  Motion of the connecting rods.  Sideways motion of balance shaft weights.

The second of these is the important thought for secondary balance. There are two important management mechanisms for secondary balance—matching the phasing of

Fig.1 pistons along the crank, as a result their second order contributions get cancelled, and the use of balance shafts that The above picture shows a rigid rotor rotating at a run at twice engine speed, and thus will provide a force for relentless angular velocity ω rad/s whereas four masses are balancing. No commonly used engine configuration is placed in same transverse plane however at totally different perfectly balanced in respect of secondary excitation. angular and radial positions. However by practicing sure definitions for secondary balance, particular configurations will be correctly claimed to be For static balance, moderately balanced no matter these restricted senses.

Σ mrcosθ = m1r1cosθ1 + m2r2cosθ2 + m3r3cosθ3 + 5. Theoretical Analysis Of Balancing Of Multi-Cylinder m4r4cosθ4 = 0 and Σ mrsinθ = m1r1sinθ1 + m2r2sinθ2 + V-Engines: A V engine is a common configuration for an IC m3r3sinθ3 + m4r4sinθ4 = 0 engine. The axial planes where the two sets of pistons reciprocate intersect at the crankshaft axis and form a V of 2. Dynamic Balancing: When several masses rotate in angle β. In case of automotive installations, V-6 and V-8 several planes, the unbalanced centrifugal forces form an engines are used generally in which β is either 60˚ or 90˚. The unbalanced couple .A system of rotating masses is claimed to 90˚ V-angle is most preferred and generally used. The V

International Journal of Scientific Engineering and Technology Research Volume.04, IssueNo.57, December-2015, Pages: 12325-12340 Computer-Aided Analysis of Balancing Of Multi-Cylinder Radial and V- Engines configuration generally helps in reducing the overall engine Σ (aisin2ϕi) = a1*(sin2ϕ1) + a2*(sin2ϕ2) + a3*(sin2ϕ3) length, height and weight compared to a counterpart inline +………. + an-1*(sin2ϕn-1) + an*(sin2ϕn) Where a1= (n-1)/2 configuration. The most common configuration used is V2, , a2 = a1-1,a3 = a2 – 1,...... an = an-1 – 1 V4, V6, V8, V10 and V12. The forces and couples for different cylinder numbers and firing order are analyzed If ―l‖ represents the distance between any two consecutive below theoretically. cylinders then, Primary couple along the central axis (CP)

V Engine with 2n cylinders =mrω2l*cosα[{cos(θ+α)+cos(θ-α)} Σ (aicosϕi) – {sin(θ+α)+sin(θ-α)} Σ (aisinϕi)] Secondary couple along the No. Of cranks: n central axis (CS)= mrω2l/n*cosα[{cos2(θ+α)+cos2(θ- α)}Σ(aicos2ϕi)–{sin2(θ+α)+sin2(θ-α)}Σ(aisin2ϕi)] Angle made by the first crank with the axis = θ 7. V-2 Engine Analysis: V angle = β= 2α

Angular Positions of cranks on the crankshaft

ϕ1 = 0° ϕ2 = ϕ1 ϕ3 = ϕ2

: : : : : : Φn-1 = ϕn-2 Φn = ϕn-1

6. Shaking Forces (cosϕi) = (cosϕ1) + cosϕ2 + cosϕ3 +………. + cosϕn-1 + Fig.2. cosϕn sinϕi = sinϕ1 + sinϕ2 + sinϕ3 +……….+ sinϕn-1 + sinϕn ϕ1 = 0° cos2ϕi = cos2ϕ1 + cos2ϕ2 + cos2ϕ3 +……….cos2ϕn-1 + cos2ϕn hence Σ(cosϕi) = 1= Σ cos2ϕi , Σ sin2ϕi = 0 = Σ sinϕi FPx = sin2ϕi = sin2ϕ1 + sin2ϕ2 + sin2ϕ3 +………. + sin2ϕn-1 + mrω2 cosα [{cos(θ+α)+ cos(θ-α)} ] sin2ϕn FPy = mrω2 sinα [{cos(θ-α)- cos(θ+α)}] Primary force along x- axis =FPx , FP = ((FPx)2+ (FPY)2)½

Primary force along y- axis = FPy Total primary force = FP FSx= mrω2 /ncos2α [{cos2(θ+α)+ cos2(θ-α)} ] FSy= mrω2 /nsin2α [{cos2(θ-α)- cos2(θ+α)} ] Similarly secondary force along x& y axis are FSx and FSY respectively Total secondary force = FS FS = ((FSx)2+ (FSY)2)½ CP =CS = 0 FPx = mrω2 cosα [{cos(θ+α)+ cos(θ-α)} Σ (cosϕi) – {sin( θ+α)+ sin( θ-α)}Σ (sinϕi)] FPy = mrω2 sinα [{cos(θ-α)- 8. V-6 Engine Analysis cos(θ+α)} Σ (cosϕi) – {sin( θ-α)- sin( θ+α)}Σ (sinϕi)] And, FP = ((FPx)2+ (FPY)2)½

FSx= mrω2 /ncos2α [{cos2(θ+α)+ cos2(θ-α)} Σ (cos2ϕi) – {sin2( θ+α)+ sin2( θ-α)}Σ (sin2ϕi)] FSy= mrω2 /nsin2α [{cos2(θ-α)- cos2(θ+α)} Σ (cos2ϕi) – {sin2( θ-α)- sin2( θ+α)}Σ (sin2ϕi)] FS = ((FSx)2+ (FSY)2)½

6.SHAKING COUPLES/MOMENTS (aicosϕi) = a1*(cosϕ1) + a2*(cosϕ2) + a3*(cosϕ3) +………. + an-1*(cosϕn-1) + an*(cosϕn) Fig.3. (aisinϕi) = a1*(sinϕ1) + a2*(sinϕ2) + a3*(sinϕ3) +………. + ϕ1 = 0° ϕ2 = 1200 an-1*(sinϕn-1) + an*(sinϕn) ϕ3 = 2400 (aicos2ϕi) = a1*(cos2ϕ1) + a2*(cos2ϕ2) + a3*(cos2ϕ3) +………. + an-1*(cos2ϕn-1) + an*(cos2ϕn) 9. Shaking Forces: cosϕi = cos0° + cos240° + cos120° = 0

International Journal of Scientific Engineering and Technology Research Volume.04, IssueNo.57, December-2015, Pages: 12325-12340 D. AMRUTHA VIJAY, SNEHALATHA. P sinϕi = sin0° + sin240° + sin120°= 0 Secondary couple along the central axis (CS)= 0 cos2ϕi = cos0° + cos480° + cos240° = 0 sin2ϕi = sin0° + sin480° + sin240°= 0 Hence FPx = FPy = The previous theoretical analysis can also be done for FSx = FSy = 0 varied engines with increasing range of cylinders like V-14, V-16 and thus on. but calculations will be much more tedious, 10. Shaking Couples: thus we are going to choose computer-aided analysis of (aicosϕi) = 1.5 balancing of multi-cylinder V engines. A program has been (aisinϕi) = 0.866 developed in C language for the specified analysis. The (aicos2ϕi) = 1.5 program can evaluate all forces and couples working on the (aisin2ϕi) = -0.866 engine for various range of cylinders and different firing orders. Primary couple along the central axis (CP) 14. C Program For Dynamic Analysis Of Multi-Cylinder =mrω2l*cosα[1.5{cos(θ+α)+cos(θ-α)} – V-Engines 0.866{sin(θ+α)+sin(θ-α)} ] #include #include #include Secondary couple along the central axis (CS)= mrω2l/n*cosα[1.5{cos2(θ+α)+cos2(θ- const double Pi=3.14159265; int main() α)}+0.866{sin2(θ+α)+sin2(θ-α)}] { 11. V-10 Engine Analysis int k; float

i,j,m,ls,R,L,N,b,o,l,n,w,a,O[10],D[10],A[10],AOC,AOCC,AO S,AOSS,TC,TS,TTC,TTS,x,y,fpx,fpy,f

sx,fsy,fp,fs,TUF,cp,cs;

printf("\nEnter Total no of cylinders\t:"); scanf("%f",&j);

printf("\nEnter mass of each cylinder\t:"); scanf("%f",&m);

printf("\nEnter length of each stroke\t:"); scanf("%f",&ls);

printf("\nEnter length of connecting rod\t:"); scanf("%f",&L);

printf("\nEnter RPM of Engine\t:"); scanf("%f",&N);

Fig.5. printf("\nEnter V Angle\t:"); scanf("%f",&b);

ϕ1 = 0° ϕ2 = 720 ϕ3 = 1440 printf("\nEnter Angle between 1st cyl and V-axis\t:"); scanf("%f",&o); ϕ4 = 2880 ϕ5 = 2160 printf("\nEnter Distance betwwen two consecutiv cyl 12. Shaking Forces bank\t:"); scanf("%f",&l); cosϕi = cos0° + cos216° + cos144° + cos72° + cos288° = 0 sinϕi = sin0° + sin216° + sin144° + sin72° + sin288° = 0 i=j/2; cos2ϕi = cos0° + cos432° + cos288° + cos144° + cos576°= 0 sin2ϕi = sin0° + sin432° + sin288° + sin144° + sin576° = 0 R=ls/2; Hence FPx = FPy = FSx = FSy = 0 n=L/R; 13. Shaking Couples (aicosϕi) = 3 b=b*(Pi/180); (aisinϕi) = 1 (aicos2ϕi) = 0 a=b/2; (aisin2ϕi) = -0 o=o*(Pi/180); Primary couple along the central axis (CP) =mrω2l*cosα[3{cos(θ+α)+cos(θ-α)} – {sin(θ+α)+sin(θ-α)} ] x=o+a; y=o-a; O[0]=0; D[0]=0; A[0]=(i-1)/2;

AOC=A[0]*cos(O[0]); printf("\nTS= %f\t",TS); printf("\nTTC= %f\t",TTC); AOS=A[0]*sin(O[0]); printf("\nTTS= %f\t",TTS); printf("\nAOC= %f\t",AOC); AOCC=A[0]*cos(2*O[0]); printf("\nAOCC= %f\t",AOCC); printf("\nAOS= %f\t",AOS); AOSS=A[0]*sin(2*O[0]); printf("\nAOSS= %f\t",AOSS); printf("\nfpx= %f\t",fpx); w=(2*Pi*N)/60; printf("\nfpy= %f\t",fpy); printf("\nfsx= %f\t",fsx); printf("\nfsy= %f\t",fsy); printf("\nfp= %f\t",fp); printf("\nfs= TC=cos(O[0]); %f\t",fs); printf("\nTUF= %f\t",TUF); printf("\ncp= %f\t",cp); printf("\ncs= %f\t",cs); TS=sin(O[0]); getch(); return 0; TTC=cos(2*O[0]); } TTS=sin(2*O[0]); 15. Theoretical Analysis Of Balancing Of Multi-Cylinder for(k=1;k

International Journal of Scientific Engineering and Technology Research Volume.04, IssueNo.57, December-2015, Pages: 12325-12340 D. AMRUTHA VIJAY, SNEHALATHA. P ΦSr1 = 00, ΦSr2 = 5400 Hence primary force due to direct crank (FPd) = (m/2) *r*ω2*( Σ cosΦPdi) Primary force due to reverse crank ( FPr) Σcos ΦSdi = cos ΦSd1 +cos ΦSd2+cos = (m/2)*r*ω2*( Σ cosΦPri) Hence primary unbalance force ΦSd3+...... cos ΦSdn Σcos ΦSri = cos ΦSr1+ cos (FP) = FPd +FPr ΦSr2+cos ΦSr3+...... cos ΦSrn

18. Secondary Unbalance Force Secondary force due to direct crank (FSd) = For direct crank; (m/2)*(r/n)*ω2* Σcos ΦSdi Secondary force due to reverse crank(FSr) = (m/2)*(r/n)*ω2* Σcos ΦSri Total secondary ΦSdi = -θi, , i.e. ΦSd1 = 00 , ΦSd2 = -3600/n , ΦSd3 = -2* unbalance force(FS) = FSd+FSr Total unbalanced force(F) = 3600/n , ...... ΦSdn = - (n-1)*3600/n For reverse FP+FS Total unbalanced couple = 0. crank; R-4 Engine Analysis: ΦSri =3*θi , i.e. ΦSr1 = 00, ΦSr2 = 3*3600/n , ΦSr3 = 3*(2*3600/n) ,...... ΦSrn = 3*{(n-1)*3600/n} Σcos ΦSdi = cos ΦSd1 +cos ΦSd2+cos ΦSd3+...... cos ΦSdn Σcos ΦSri = cos ΦSr1+ cos ΦSr2+cos ΦSr3+...... cos ΦSrn Secondary force due to direct crank (FSd) = (m/2)*(r/n)*ω2*

Σcos ΦSdi Secondary force due to reverse crank(FSr) = Fig.7. (m/2)*(r/n)*ω2* Σcos ΦSri Total secondary unbalance force(FS) = FSd+FSr Angle between each stroke axis = 3600/4 = 900 Hence θ1=00, θ2= 1200, θ3 = 2400 Total unbalanced force(F) = FP+FS Total unbalanced couple = 0 Primary Unbalance Force: For Direct crank ; 19. R-2 Engine Analysis ΦPd1 = ΦPd2 = ΦPd3 = ΦPd4 = 00

For reverse crank;

ΦPr1 = 00, ΦPr2 = 1800 , ΦPr3 = 3600 ΦPr4 = 5400 cosΦPdi = 4 cos ΦPri = cos(00) + cos(1800) + cos (3600) +cos(5400) = 0 Fig.6. Hence primary force due to direct crank (FPd) = 4*(m/2) *r*ω2 Angle between each stroke axis = 3600/2 = 1800 Primary force due to reverse crank ( FPr) = 0 Hence θ1=00, θ2= 1800 Hence primary unbalance force (FP) = FPd

20. Primary Unbalance Force Secondary Unbalance Force: For Direct crank ; For direct crank; ΦPd1 = ΦPd2 = 00 ΦSdi = -θi, , i.e. ΦSd1 = 00 , ΦSd2 = -900 , ΦSd3 = -1800 , ΦSd4 = -2700 For reverse crank; ΦPr1 = 00, ΦPr2 =3600 For reverse crank; cosΦPdi = 2 ΦSri =3*θi , i.e. ΦSr1 = 00, ΦSr2 = 2700 , ΦSr3 = 5400 , cos ΦPri = cos(00) + cos(3600)=2 ΦSr4 = 8100

Hence primary force due to direct crank (FPd) = (m/2) Σcos ΦSdi = cos ΦSd1 +cos ΦSd2+cos ΦSd3 + cos ΦSd4 = 0 *r*ω2*2 = m*r* ω2 Primary force due to reverse crank ( FPr) Σcos ΦSri = cos ΦSr1+ cos ΦSr2+cos ΦSr3 + cos ΦSr4 = 0 = (m/2)*r*ω2*2 = m*r* ω2 Hence primary unbalance force (FP) = 2* m*r* ω2 Secondary force due to direct crank (FSd) = (m/2)*(r/n)*ω2* Σcos ΦSdi = 0 Secondary force due to reverse crank(FSr) = 0 Secondary Unbalance Force For direct crank; Total secondary unbalance force(FS) = 0 Total ΦSdi = -θi, , i.e. ΦSd1 = 00 , ΦSd2 = -1800 unbalanced force(F) = FP+FS = Fp Total unbalanced couple = 0 For reverse crank; R-6 Engine Analysis:

Primary Unbalance Force: For Direct crank ; ΦPd1 = ΦPd2 = ΦPd3 = ΦPd4 = ΦPd5 = ΦPd6 = ΦPd7= ΦPd8= 00

Fig.8. For reverse crank; ΦPr1 = 00, ΦPr2 =900 , ΦPr3 = 1800, ΦPr4 = 2700 , ΦPr5 Angle between each stroke axis = 3600/6 = 600 =3600 , ΦPr6 = 4500, ΦPr7 = 5400 , ΦPr8 =6300

Hence θ1=00, θ2= 600 , θ3 = 1200 , θ4 = 1800 ,θ5 = 2400 , cosΦPdi = 8 θ6 = 3000 cos ΦPri = 0 Primary Unbalance Force: For Direct crank ; Hence primary force due to direct crank (FPd) = 8*(m/2) ΦPd1 = ΦPd2 = ΦPd3 = ΦPd4 = ΦPd5 = ΦPd6 = 00 *r*ω2 Primary force due to reverse crank ( FPr) = 0 For reverse crank; Hence primary unbalance force (FP) = FPd ΦPr1 = 00, ΦPr2 =1200 , ΦPr3 = 2400, ΦPr4 = 3600 , ΦPr5 = 4800 , ΦPr6 = 6000 Secondary Unbalance Force: cosΦPdi = 6 For direct crank; cos ΦPri = o ΦSd1 = 00, ΦSd2 =-450,ΦSd3 =-900 ,ΦSd4 =-1350 ,ΦSd5=- 1800,ΦSd6=-2250, ΦSd7=-2700, ΦSd8=-3150 For reverse Hence primary force due to direct crank (FPd) = 6*(m/2) crank; *r*ω2 ΦSr1 = Primary force due to reverse crank ( FPr) = 0 00,ΦSr2=1350,ΦSr3=2700,ΦSr4=4050,ΦSr5=5400,ΦSr6=67 Hence primary unbalance force (FP) = FPd 50, ΦSr7=8100, ΦSr8=9450 Σcos ΦSdi = 0 Σcos ΦSri = 0 Secondary Unbalance Force: Secondary force due to direct crank (FSd) = 0 Secondary For direct crank; force due to reverse crank(FSr) = 0 Total secondary ΦSdi = -θi, , i.e. ΦSd1 = 00 , ΦSd2 = -600, ΦSd3 = -1200 , unbalance force(FS) = 0 Total unbalanced force(F) = FP ΦSd4 = -1800 , ΦSd5 = - 2400 , ΦSd6 = - 3000 Total unbalanced couple = 0

For reverse crank; R-10 Engine Analysis: ΦSri =3*θi , i.e. ΦSr1 = 00, ΦSr2 = 1800 , ΦSr3 = 3600 , ΦSr4 = 5400, ΦSr5 = 7200 , ΦSr6 = 9000

Σcos ΦSdi = 0 Σcos ΦSri = 0 Secondary force due to direct crank (FSd) = 0 Secondary force due to reverse crank(FSr) = 0 Total secondary unbalance force(FS) = 0 Total unbalanced force(F) = FP

Total unbalanced couple = 0 Fig.10.

R-8 Engine Analysis: Angle between each stroke axis = 3600/10 = 360 Hence θ1=00, θ2= 360 , θ3=720,θ4= 1080, θ5 = 1440 ,θ6 = 1800 , θ7 = 2160, θ8 = 2520, θ9=2880, θ10=3240

Primary Unbalance Force: For Direct crank ; ΦPd1 = ΦPd2 = ΦPd3 = ΦPd4 = ΦPd5 = ΦPd6 = ΦPd7= ΦPd8= ΦPd9= ΦPd10=0 Fig.9. For reverse crank; Angle between each stroke axis = 3600/8 = 450 ΦPr1 = 00, ΦPr2 =720 ,ΦPr3= 1440,ΦPr4=2160,ΦPr5=2880, ΦPr6=3600,ΦPr7=4320,ΦPr8=5040,ΦPr9=5760,ΦPr10=6480

International Journal of Scientific Engineering and Technology Research Volume.04, IssueNo.57, December-2015, Pages: 12325-12340 D. AMRUTHA VIJAY, SNEHALATHA. P cosΦPdi = 10 Hence primary unbalance force (FP) = FPd cos ΦPri =0 Secondary Unbalance Force: For direct crank; Hence primary force due to direct crank (FPd) = 10*(m/2) ΦSd1=00,ΦSd2=-300,ΦSd3=-600,ΦSd4=-900,ΦSd5=- *r*ω2 1200,ΦSd6=-1500,ΦSd7=-1800,ΦSd8=- Primary force due to reverse crank ( FPr) = 0 Hence primary unbalance force (FP) = FPd 2100,ΦSd9=-2400, ΦSd10=-2700 , ΦSd11=-3000 , ΦSd12=- 3300 For reverse crank; Secondary Unbalance Force: For direct crank; ΦSd1=00,ΦSd2=-360,ΦSd3=-720,ΦSd4=- ΦSr1= 1080,ΦSd5=-1440,ΦSd6=-1800,ΦSd7=-2160,ΦSd8=- 00,ΦSr2=900,ΦSr3=1800,ΦSr4=2700,ΦSr5=3600,ΦSr6=450 2520,ΦSd9=-2880, 0,ΦSr7=5400,ΦSr8=6300, ΦSr9=7200, ΦSr10=8100 , ΦSd10=-3240 ΦSr11=9000, ΦSr12=9900 For reverse crank; ΦSr1= Σcos ΦSdi = 0 Σcos ΦSri = 0 00,ΦSr2=1080,ΦSr3=2160,ΦSr4=3240,ΦSr5=4320,ΦSr6=54 00,ΦSr7=6480,ΦSr8=7560, ΦSr9=8640, ΦSr10=9720 Secondary force due to direct crank (FSd) = 0 Secondary Σcos ΦSdi = 0 Σcos ΦSri = 0 force due to reverse crank(FSr) = 0 Total secondary Secondary force due to direct crank (FSd) = 0 Secondary unbalance force(FS) = 0 Total unbalanced force(F) = FP force due to reverse crank(FSr) = 0 Total secondary unbalance force(FS) = 0 Total unbalanced force(F) = FP Total unbalanced couple = 0

Total unbalanced couple = 0 The above analytical analysis can be done for radial engines with more no. Of cylinders but calculation will be R-12 Engine Analysis: more hence computer aided analysis is preferred to minimize time and calculation. A c programme is generated for the above purpose.

21. C Program For Dynamic Analysis Of Multi-Cylinder Radial Engines:

#include

Fig.11. #include

Angle between each stroke axis = 3600/12 = 300 const double PI = 3.14159265; int main() Hence θ1=00, θ2= 300 , θ3=600,θ4= 900, θ5 = 1200 ,θ6 = 1500 , θ7 = 1800, θ8 = 2100, θ9=2400, θ10=2700, θ11=3000, { θ12=3300 float Primary Unbalance Force: i,a,n,r,m,w,l,N,angle,ang1[20],ang2[20],s=0,s1=0,ang3[20],s2 For Direct crank ; =0,fp,fpr,ang[20],fpd,fsr,fsd,f,fs; ΦPd1 = ΦPd2 = ΦPd3 = ΦPd4 = ΦPd5 = ΦPd6 = ΦPd7= ΦPd8= ΦPd9= ΦPd10= ΦPd11 = ΦPd12 =0 printf("enter no of cylinders :");scanf("%f",&i);printf("\n");

For reverse crank; printf("enter the length of each stroke(in metre) ΦPr1 = 00, ΦPr2 =600 ,ΦPr3= :");scanf("%f",&a);printf("\n"); printf("enter mass of each 1200,ΦPr4=1800,ΦPr5=2400,ΦPr6=3000,ΦPr7=3600,ΦPr8= cylinder (in kg) :");scanf("%f",&m);printf("\n"); printf("enter 4200,ΦPr9=4800, ΦPr10=5400 , length of connecting rod(in metre) :"); scanf("%f",&l); ΦPr11=6000, ΦPr12 = 6600 printf("\n"); printf("enter rpm of the engine :"); scanf("%f", cosΦPdi = 12 &N); printf("\n");r=a/2;n=l/r;printf("%f %f\n",r,n); cos ΦPri =0 Hence primary force due to direct crank (FPd) = 12*(m/2) w=(2*PI*N)/60;printf("%f\n",w); *r*ω2 angle=360/i;ang[0]=0;ang1[0]=0;ang2[0]=0;//printf("%f\n",a Primary force due to reverse crank ( FPr) = 0 ngle); for(int j=1;j<=i;j++)

International Journal of Scientific Engineering and Technology Research Volume.04, IssueNo.57, December-2015, Pages: 12325-12340 Computer-Aided Analysis of Balancing Of Multi-Cylinder Radial and V- Engines Distance between each row ―l‖ { ang[j]=ang[j-1]+angle;ang2[j]=-ang[j]; ang3[j]=3*ang[j]; Referring analysis of radial engine of chapter 5 ang1[j]=2*ang[j];s=s+cos(ang1[j- Total unbalanced primary force of an individual R-3 engine = 1]*PI/180);s1=s1+cos(ang2[j- 3*(m/2)*rω2

1]*PI/180);s2=s2+cos(ang3[j-1]*PI/180); Total unbalanced secondary force for an individual R-3 engine = 3*(m/2)*(r/n)ω2 Hence total Primary force for 2- printf("%f %f\n",ang[j-1],cos(ang[j-1]*PI/180)); Row R-3 engine(FP) = 2*3*(m/2)*rω2

} Total secondary force for 2- Row R-3 engine (FS) = 2*= 3*(m/2)*(r/n)ω2 Total unbalanced force (F) = FP+FS //printf("%f %f \n",ang[j],cos(ang[j]*PI/180)); Total primary couple (CP) = 0 Total secondary couple (CS) = //s=s+cos(ang1[11]*PI/180);s1=s1+cos(ang2[11]*PI/180);s2= 0 s2+cos(ang3[11]*PI/180); printf("%f %f %f\n",s,s1,s2); B. 4- Row R-3 Engine Analysis fpd=i*.5*m*r*pow(w,2);printf("total primary force for direct crank %f\n",fpd); fpr=m*r*pow(w,2)*s/2;printf("total primary force for reverse crank %f\n",fpr); fp=fpd+fpr;printf("total primary force %f\n",fp); fsd=m*(r/n)*pow(w,2)*s1/2;printf("total secondary force for direct crank %f\n",fsd); fsr=m*(r/n)*pow(w,2)*s2/2;printf("total secondary force for reverse crank %f\n",fsr); fs=fsd+fsr;printf("total secondary force %f\n",fs);

Fig.13. f=fp+fs;printf("total unbalanced force %f\n",f);

getch(); return 0; Distance between each row ―l‖

} Referring analysis of radial engine of chapter 5

Total unbalanced primary force of an individual R-3 III. THEORETICAL ANALYSIS OF BALANCING OF engine = 3*(m/2)*rω2 MULTI-CYLINDER MULTIPLE ROW RADIAL

ENGINES Total unbalanced secondary force for an individual R-3 In aero engines generally radial engines are connected engine = 3*(m/2)*(r/n)ω2 Hence total Primary force for 4- axially with a common crank shaft as a result whole unit Row R-3 engine(FP) = 4*3*(m/2)*rω2 works as one configuration. In general practice two rows of radial engines are used. Maximum four rows are practically Total secondary force for 4- Row R-3 engine (FS) = used. 4*3*(m/2)*(r/n)ω2 Total unbalanced force (F) = FP+FS A. 2- Row R-3 Engine Total primary couple (CP) = 0 Total secondary couple (CS) = 0

C. 2- Row R-4 Engine Analysis

Fig.12. Fig.14.

International Journal of Scientific Engineering and Technology Research Volume.04, IssueNo.57, December-2015, Pages: 12325-12340 D. AMRUTHA VIJAY, SNEHALATHA. P

Distance between each row ―l‖ ΦPd1 = ΦPd2 = ΦPd3 = ΦPd4 =...... = ΦPdn = 00 Referring analysis of radial engine of chapter 5 For reverse crank; Total unbalanced primary force of an individual R-4 engine = 4*(m/2)*rω2 Total unbalanced secondary force for ΦPr1 = 00, ΦPr2 =2* 3600/n , ΦPr3 = 2*(2* 3600/n), an individual R-4 engine = 0 ...... ΦPrn = 2*{(n-1)*3600/n} Hence ΦPri = 2*θi Hence total Primary force for 2-Row R-4 engine(FP) = cosΦPdi = n 2*4*(m/2)*rω2 Total secondary force for 2- Row R-4 engine (FS) = 0 cos ΦPri = cos(ΦPr1) + cos(ΦPr2) + cos (ΦPr3) +...... cos(ΦPrn) Total unbalanced force (F) = FP+FS Total primary couple Hence primary force due to direct crank (FPd) = (m/2) (CP) = 0 Total secondary couple (CS) = 0 *r*ω2*( Σ cosΦPdi) Primary force due to reverse crank ( FPr) = (m/2)*r*ω2*( Σ cosΦPri) Hence primary unbalance force D. 4-Row R-4 Engine Analysis (FP) = FPd +FPr

Secondary Unbalance Force: For direct crank; ΦSdi = -θi, , i.e. ΦSd1 = 00 , ΦSd2 = -3600/n , ΦSd3 = -2* 3600/n , ...... ΦSdn = - (n-1)*3600/n For reverse crank;

ΦSri =3*θi , i.e. ΦSr1 = 00, ΦSr2 = 3*3600/n , ΦSr3 = 3*(2*3600/n) ,...... ΦSrn = 3*{(n-1)*3600/n} Σcos ΦSdi = cos ΦSd1 +cos ΦSd2+cos ΦSd3+...... cos ΦSdn Σcos ΦSri = cos ΦSr1+ cos ΦSr2+cos ΦSr3+...... cos ΦSrn Fig.15. Secondary force due to direct crank (FSd) = (m/2)*(r/n)*ω2* Σcos ΦSdi Secondary force due to reverse crank(FSr) = Distance between each row ―l‖ (m/2)*(r/n)*ω2* Σcos ΦSri Total secondary unbalance force(FS) = FSd+FSr Referring analysis of radial engine of chapter 5 Hence total Primary force for j-Row R-n engine(FP) = Total unbalanced primary force of an individual R-4 j*[(m/2) *r*ω2*( Σ cosΦPdi) + engine = 4*(m/2)*rω2 Total unbalanced secondary force for (m/2)*r*ω2*( Σ cosΦPri)] an individual R-4 engine = 0 Total secondary force for j- Row R-n engine (FS) = j* [(m/2)*(r/n)*ω2* Σcos ΦSdi + Hence total Primary force for 4-Row R-4 engine(FP) = (m/2)*(r/n)*ω2* Σcos ΦSri) 4*4*(m/2)*rω2 Total secondary force for 4- Row R-4 engine (FS) = 0 Total unbalanced force (F) = FP+FS Total primary couple (CP) = 0 Total secondary couple (CS) = 0 Total unbalanced force (F) = FP+FS Total primary couple (CP) = 0 Total secondary couple (CS) = 0 The above analytical analysis can be done for combined row radial engines with more no. Of cylinders but calculation E. j- Row R-n Engine Analysis will be more hence computer aided analysis is preferred for time and calculation minimization. A c programme is Referring to chapter 5 generated for the above purpose.

Angle between each stroke axis = 3600/n E. C Program For Dynamic Analysis Of Multi-Cylinder Multiple Row Radial Engines Hence θ1=00, θ2= 3600/n , θ3 = 2* 3600/n , θ4 = 3* 3600/n ,...... θn = (n-1)* 3600/n include

Primary Unbalance Force: include

For Direct crank ; include

International Journal of Scientific Engineering and Technology Research Volume.04, IssueNo.57, December-2015, Pages: 12325-12340 Computer-Aided Analysis of Balancing Of Multi-Cylinder Radial and V- Engines fs=fsd+fsr;printf("total secondary force %f\n",fs); const double PI = 3.14159265; int main() f=fp+fs;printf("total unbalanced force %f\n",f);

{ getch(); return 0; float } i,j,k,a,n,R,m,w,l,L,N,angle,ang1[20],ang2[20],s=0,s1=0,ang3[ 20],s2=0,fp,fpr,ang[20],fpd,fsr, fsd,f,fs; IV. RESULTS AND DISCUSSION  It is assumed that the mass of the cylinder is 2.4 kg printf("enter no of cylinders :");scanf("%f",&i);printf("\n"); RPM of the engine is 2000. printf("enter no of rows :");scanf("%f",&j);printf("\n");  Length of each stroke 0.16 m Length of connecting rod 0.24 m Vee angle 600. printf("enter mass of each cylinder (in kg)  Angle between 1st cylinder and Vee – axis 00 :");scanf("%f",&m);printf("\n"); printf("enter the length of Distance between two consecutive cylinders 0.1 m. each stroke(in metre) :");scanf("%f",&a);printf("\n");  Now varying no. Of cylinders present in the printf("enter length of connecting rod(in metre) combination unbalanced force and couple was :");scanf("%f",&L);printf("\n"); printf("enter rpm of the calculated. engine :");scanf("%f",&N);printf("\n"); For V-2 Engine : printf(" enter the distance between two consecutive bank of radial cylinder:"); scanf("%f",&l); printf ("\n");

R=a/2;n=L/R;printf("%f %f\n",R,N); w=(2*PI*N)/60;printf("%f\n",w); k= i/j; angle=360/k;ang[0]=0;ang1[0]=0;ang2[0]=0;//printf("%f\n",a ngle); for(int p=1;p<=k;p++)

{ ang[p]=ang[p-1]+angle;ang2[p]=-ang[p]; ang3[p]=3*ang[p]; ang1[p]=2*ang[p];s=s+cos(ang1[p- For V-4 Engine: 1]*PI/180);s1=s1+cos(ang2[p-

1]*PI/180);s2=s2+cos(ang3[p-1]*PI/180); printf("%f %f\n",ang[p-1],cos(ang[p-1]*PI/180));

}

//printf("%f %f \n",ang[p],cos(ang[p]*PI/180));

//s=s+cos(ang1[11]*PI/180);s1=s1+cos(ang2[11]*PI/180);s2= s2+cos(ang3[11]*PI/180); printf("%f %f %f\n",s,s1,s2); For V-6 Engine: fpd=i*.5*m*R*pow(w,2);printf("total primary force for direct crank %f\n",fpd); fpr=j*m*R*pow(w,2)*s/2;printf("total primary force for reverse crank %f\n",fpr); fp=fpd+fpr;printf("total primary force %f\n",fp); fsd=j*m*R*pow(w,2)*s1*0.5/n;printf("total secondary force for direct crank

%f\n",fsd); fsr=j*m*R*pow(w,2)*s2*0.5/n;printf("total secondary force for reverse crank %f\n",fsr);

International Journal of Scientific Engineering and Technology Research Volume.04, IssueNo.57, December-2015, Pages: 12325-12340 D. AMRUTHA VIJAY, SNEHALATHA. P For V-8 Engine:  Values considered for radial engine: Mass of each cylinder : 2,4 kg RPM of the engine 2000.  Length of each stroke 0.16m Length of connecting rod 0.24m.  Unbalance forces are calculated for varying no. Of cylinders.

For R-2 Engine:

For R-3 Engine:

For V-10 Engine:

For R-4 Engine: For V-12 Engine:

For R-10 Engine: For R-6 Engine:

For R-11 Engine: For R-7 Engine:

For R-12 Engine: For R-8 Engine:

International Journal of Scientific Engineering and Technology Research Volume.04, IssueNo.57, December-2015, Pages: 12325-12340 D. AMRUTHA VIJAY, SNEHALATHA. P  Values taken for multiple rows combined radial 2 Rows R-4 Engine: engine Mass of each cylinder 2.4 kg  Length of each stroke 0.16m Length of connecting rod 0.24m  Speed of the engine 2000 RPM Distance between each bank 0.1m

2- Row R-3 Engine:

3 Rows R-4 Engine:

3 Rows R-3 Engine

4 Rows R-4 Engine:

4 Rows R-3 Engine:

2 Rows R-5 Engine:

TABLE V: No. Of Cylinders = 9

TABLE VI: No. Of Cylinders = 10

2 Rows R-6 Engine:

TABLE VII: No. Of Cylinders = 12

The above data shows primary and secondary unbalanced forces are 0 for v configured engine having engines more than four. Where as in case of radial engine secondary unbalanced force and couple are 0.when R-3 engines are combined V. COMPARISONS OF UNBALANCED FORCES AND together the total unbalance force is more than the counterpart COUPLE FOR DIFFERENT CONFIGURATION radial engine having total no. Of cylinders are same. TABLE I: No. Of Cylinders = 2 VI. CONCLUSION On comparing radial engines and multiple row radial engines it is found that the multiple row radial engines are better as they are easy to manufacture. Higher no. Of cylinders can be achieved. It is highly applicable where higher power is required. it is generally used in aircraft engine. TABLE II: No. Of Cylinders = 4 VII. REFERENCES [1] Vigen H. Arakelian and M. R. Smith, Shaking Force and Shaking Moment Balancing of Mechanisms: A Historical Review with New Examples Journal of Mechanical Design, MARCH 2005, Vol. 127, pp 334-339. [2]Esat , H. Bahai, A theory of complete force and moment

balancing of planer linkage mechanisms, Mechanism and TABLE III: No. Of Cylinders = 6 Machine Theory 34 (1999) 903-922. [3]V. Arakelian and N. Makhsudyan, Generalized Lanchester Balancer , Mechanics research communications(2010), doi: 10.1016/j.mechrescom.2010.08.2003. [4]Floyd A. Wyczalek , Generalized Balance of Inline, Vee and Opposed Piston Engines, Paper Number: 880418, DOI: 10.4271/880418, Date Published: 1988-02-01.

International Journal of Scientific Engineering and Technology Research Volume.04, IssueNo.57, December-2015, Pages: 12325-12340 D. AMRUTHA VIJAY, SNEHALATHA. P [5]W Harry Close, Wieslaw Szydlowski and Christopher Downton, Perfect Engine Balance Via Collins Scotch Yoke Technology, SAE International, 941039 ,March 1994. [6]H. D. Desai, Computer Aided Kinematic and Dynamic Analysis of a Horizontal Slider Crank Mechanism Used For Single-Cylinder Four Stroke Internal Combustion Engine, Proceedings of the World Congress on Engineering 2009 Vol II WCE 2009, July 1 - 3, 2009, London, U.K. [7]A.coppens, - University Of Louvain, Belgium, Improved Formula for Computing Counterweights of Single-Row and Double-Row Radial Engines, Date Published: 1934-01- 01,Paper Number: 340083 DOI: 10.4271/340083. [8]Kwon-Hee Suh - Kia Motors Corp. ,Yoon-Ki Lee - Chonnam National Univ. ,Hi-Seak Yoon - Chonnam National Univ., A Study on the Balancing of the Three-Cylinder Engine with Balance Shaft, Date Published: 2000-03-06, Paper Number: 2000-01-0601, DOI: 10.4271/2000-01-0601. [9]Mechanisms & Dynamics of machinery: Mable & Reinholtz. [10]Therory of machines :S S Rathan. [11]Theory of machines: Abdulla Shariff. [12]Theory of machines & mechanism :Ghosh & Mallick.

Author’s Profile:

D Amrutha Vijay is a student pursuing M.Tech(CAD/CAM) in Eswar College of Engineering, Narasaraopet, Guntur, India. Email id: [email protected].

P.SNEHALATHA P M.Tech (P.hD), is having 6+ years of experience in the field of teaching in various Engineering Colleges and PG colleges. At present she is working as Asst.Prof. in Eswar College of Engineering, Narasaraopet, Guntur, India. She published 1 international journal i.e SCI Journal and attend 1 national conference and 2 international conferences. She also guided many B.Tech,& M.Tech projects. She attended five day workshop on ―Student Evaluation‖ conducted by NITTTR, Chennai. And one day seminar on ―teaching and learning process‖ at Stanns Engineering College, Chirala. Her interested areas are CAD, Composite Materials, Machine Drawing, Machine Tools, Production Planning And Control. Email id : [email protected].

International Journal of Scientific Engineering and Technology Research Volume.04, IssueNo.57, December-2015, Pages: 12325-12340