OTTO ENGINE DYNAMICS PETRESCU Florian Ion*, PETRESCU Relly Victoria**, GRECU Barbu*** *,**,***
Polytechnic University of Bucharest –Romania
Abstract: Otto engine dynamics are similar in almost all common internal combustion engines. We can speak so about dynamics of engines: Lenoir, Otto, and Diesel. The dynamic presented model is simple and original. The first thing necessary in the calculation of Otto engine dynamics, is to determine the inertial mass reduced at the piston. One uses then the Lagrange equation. Key words: Lagrange equation, dynamic model
1. INTRODUCTION The first thing necessary in the calculation of Otto engine dynamics, is to determine the inertial mass reduced at the piston (1).
J 2 cos 2 r2 J M M * mt mbA 2 12 22 s' s' s' cos 2 J1 J2 1 2 2 2 M mt [(mbA 2 ) (1 sin ) 2 cos ] 2 r l sin (cos cos ) 2 m (1 2 sin 2 ) m2 cos 2 M mt 1 2 sin (cos cos ) 2
(1)
Then it derives the reduced mass to the crank position angle (2). Were used for piston the next kinematics parameters (4). Lagrange equation is written in the form (3).
2 cos (2 m1 m2 ) dM cos sin ( M mt ) (2) ( ) d sin cos sin (cos cos ) 2 1 dM 2 M 2 x' ' x ' k ( s x ) Fp 2 d
s r cos l cos l r sin (cos cos ) s ' cos r cos(2 ) r 3 sin 2 cos 2 s ' ' r cos cos cos 3
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(2)
(3)
(4)
2. DYNAMIC EQUATIONS The dynamic equation of motion of the piston, obtained by integrating the Lagrange equation (3), takes the form 5.
k cos c3 c4 cos 2 cos (cos cos ) k mt
x s3
(5)
Dynamic reduced velocity (6) and dynamic reduced acceleration (7) are obtained by derivation:
x' s'3
k sin c3 c4 sin 2 cos (cos cos ) k mt
(6)
x' ' s' '3
k cos c3 c4 cos 2 cos (cos cos ) k mt
(7)
Angular velocity * is obtained through kinetic energy conservation (8-12).
1 * 1 J *2 J D* D2 2 2 2 2 D m D m (cos ) m cos m (1 sin 2 ) m (1 2 sin 2 ) J J1 mbA r mt s' *
2
2
J J1 mbA r mt x' * D
*
2
(8) (9) (10)
2
(11)
J 1 mbA r 2 mt x' 2 n (1 2 sin 2 ) J 1 mbA r 2 mt s' 2 30
(12)
Dynamic velocity (13) and kinematics velocity (14) are written:
x x' *
s s' m s'
(13)
n
(14)
30
Dynamic acceleration (15) and kinematics acceleration (16) are written:
x x' ' *2
s s' 'm2 s' '
(15)
n 2
2
900
(16)
3. NOTATIONS In the picture number 1 one presents the crank shaft.
k
3 E G ( Dm4 d m4 ) l p .4 D p l m .4 Dm 8r 1.6( D p Dm ) 4G(l m b) 3 96 Er 2 sin 2 ( Dm4 d m4 )[ 4 4 ] D p d p4 Dm d m4 b(2r D p Dm ) 3
(17)
The relation (17) determines the elastic constant of the crank shaft, k. For the masses one uses the notations (18); see the picture two.
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the ratio between lengths of crank and rod;
r l
m p the mass of the piston, with piston bolt and segments;
mb the mass of the rod;
Dm
dm
h
r
dp lp/2
Dp
lm
b
Fig. 1 – Crank Shaft
l'' l' mbA mb l mbB mb l mt m p mbB J1 m1 mbA 2 r J2 m2 l 2
l 'l ' ' l
mbA mbB mb (18)
The parameters c1-c4 take the forms (19):
r 2 m [ ] c1 k kg c 2 m m [kg] 2 1 2 c c c [ m] 1 2 3 c4 c1 mt [m] The moment of inertia J 1 can be determined with the relation (20).
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(19)
J1
32
{(l p 2 b) ( D p4 d p4 ) (l m 2 b) [( Dm4 d m4 ) ( Dm2 d m2 ) 8 r 2 ]} (20)
The crank length, r, and the length of the connecting-rod, l, can be seen in the kinematics schema of an Otto mechanism (fig. 2).
mp B mbB
l’’
J2
l
G l’
mb
A r
J1
mbA O
Fig. 2 – Otto mechanism kinematics schema
4. DYNAMIC ANALYSIS OF THE MECHANISM AND CONCLUSIONS When increases the mechanism dynamics is deteriorating. r=0.25 [m] l=0.3 [m] 0.8(3) For n=8000 [r/m] the mechanism is working normally (see the accelerations diagram from the picture 3): 300000 200000 100000 0 -100000
0
100
200
300
400
x2p [ms-2] s2p [ms-2]
-200000 -300000 -400000
Fig. 3 – Dynamic and kinematics accelerations; n=8000 [r/m]; 0.83
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r=0.25 [m] l=0.3 [m] 0.8(3) At n=9000 [r/m] the mechanism work abnormally (see the accelerations diagram from the picture 4): 800000 600000 400000 200000
x2p [ms-2] s2p [ms-2]
0 -200000
0
100
200
300
400
-400000 -600000
Fig. 4 – Dynamic and kinematics accelerations; n=9000 [r/m]; 0.83 r=0.25[m];l=0.3[m] For a proper operation is necessary reduction of the ratio , especially if we want to increase the engine speed (see the next diagrams). 600000 400000 200000 0 -200000
0
100
200
300
400
x2p [ms-2] s2p [ms-2]
-400000 -600000 -800000
Fig. 5 – Dynamic and kinematics accelerations; n=12000 [r/m]; r=0.25[m];l=0.6[m] 0.42 800000 600000 400000 200000 0 -200000 0
100
200
300
400
x2p [ms-2] s2p [ms-2]
-400000 -600000 -800000
Fig. 6 – Dynamic and kinematics accelerations; n=14000 [r/m]; r=0.25[m];l=0.9[m] 0.27
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We can reduce the acceleration values by reducing r and l. 100000 50000 0 0
100
200
300
400
-50000
x2p [ms-2] s2p [ms-2]
-100000 -150000 -200000
Fig. 7 – Dynamic and kinematics accelerations; n=15000 [r/m]; r=0.05[m];l=0.15[m] 0.33
80000 60000 40000 20000 0 -20000 0 -40000 -60000 -80000 -100000 -120000
100
200
300
400
x2p [ms-2] s2p [ms-2]
Fig. 8 – Dynamic and kinematics accelerations; n=50000 [r/m]; r=0.003[m];l=0.009[m] 0.33 One can reduce the acceleration values especially if we want to increase the engine speed by reducing r and l (the lengths of crank and rod).
5. REFERENCES [1] Petrescu, R.V., Petrescu, F.I., - Otto Engine Design, Acta Technica Napocensis, Series: Applied Mathematics and Mechanics, CNCSIS 118 B, ISSN 1221-5872, Vol. Ib, p. 537-540, Cluj-Napoca, 2009 [2] Petrescu, F.I., Petrescu, R.V., - V Engine Design, Acta Technica Napocensis, Series: Applied Mathematics and Mechanics, CNCSIS 118 B, ISSN 1221-5872, Vol. Ib, p. 533-536, Cluj-Napoca, 2009
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