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DYNAMICS OF RECIPROCATING ENGINES Indicator Diagrams In-Line Engines V Engines GAS FORCES Binomial Theorem: So, Since in most engines, Therefore, We can write is so small, and only the first two terms of the series may be taken with sufficient accuracy. and since and finally It is important to note that always angular velocity () is assumed constant. Therefore, acceleration of the piston may be obtained as, ------------------------------------------------------------------------------------------------------- Using Binomial theorem for the denominator of this ratio and taking only first two terms of the series, Therefore, Similarly, T12k + xi X (F14j)=0 T12 T12=-F14 x k =-P tan x k T21=P tan x k P x F12 F14 Making appropriate substitutions and neglecting higher order terms, torque delivered by the crank to the shaft is obtained EQUIVALENT MASSES • A A • m3 = G3 • G3 • B • IG3 • B • 1) Total masses should be the same: • 2) The mass center positions should be the same: • (If , ) In practice, for a connecting rod EQUIVALENT MASSES • The first two conditions is called STATIC EQIVALENCE. • 3) The mass moment of inertias should be the same: • The third condition is called DYNAMIC EQIVALENCE. • For a connecting rod of a slider-crank mechanism, our aim is to divide the total mass into two masses; one, at point A (rotating) and the other one, at point B (reciprocating). The third condition does not conform our aim , therefore it is not satisfied throughout the analysis of connecting rod. Dividing the crank mass into two parts, at O 2 and A, regarding the static equivalence conditions. +0 INERTIA FORCES Constant angular velocity assumption for crank gives the acceleration of point A as, Inertia force at A Acceleration of the piston has been found as, Inertia force of the reciprocating parts The components of total inertia force, Primary İnertia force Secondary inertia force Inertia Torque T12k + xi X (- F14j)=0 T12=F14 x k =-mBaB tan x k -mAaA T12 -mBaB x Making appropriate substitutions, neglecting higher order terms and using trigonometric identities, inertia torque exerted by the engine on the shaft is obtained , F12 Inertia torque is a periodic function, including the first three harmonics. F14=-mBaB tan BEARING LOADS IN THE SINGLE-CYLINDER ENGINE • The resultant bearing loads in a single-cylinder engine are made up of due to the following four forces: • 1) Gas force (examined at the beginning). • 2) Inertia force due to the mass of piston m4. • 3) Inertia force due to the mass of connecting rod m3B at piston-pin end. • 4) Inertia force due to the mass of connecting rod m3A at crank-pin end. • (It is assumed that crank has been balanced.) 2) Inertia force due to the mass of piston m4 F34’’ F14’’ 3) Inertia force due to the mass of connecting rod m3B at piston-pin end R F43’’’ R 4) Inertia force due to the mass of connecting rod m3A at crank-pin end A r If a counterweight of m3A is added at point C, O2 r r C r O2 FBD C A no bearing force will act at O2. Therefore shaking force due to m3A İs eliminated Superposition CRANKSHAFT TORQUE • O2 1 • • T21 F21 B MO2=0 F41 The torque delivered by the crankshaft to the load. ENGINE SHAKING FORCES (due to only reciprocating masses) for Graphically, F21x = OA’+OB’ Shaking Force: (Linear vibration in x direction) Shaking Couple: (Torsional vibration about crank center) PROBLEM Stroke=2r =90 mm d=100 mm n=4400 rpm m 3A=0.8 kg Obtain cylinder pressure for the 30% expansion case. p= 1.857 Mpa=1857000 Pa(N/m 2) P=1857000(7.854x10-3 ) = 14584.878 N r=45 mm A=d2/4=1002/4=7853.98 mm 2=7.854x10-3 m 2 =4400(2)/60=460.767 r/s m 3B=0.38 kg m 4=1.64 kg Force Calculations F41=-[(0.38+1.64)(-3578.663)+14584.878](0.1155)j F41= -849.616j N F41 F34=[1.64(-3578.663)+14584.878]i – 849.616j = 8715.87i-849.616j N