mm IFToMM ARoTMM
THE NINTH IFToMM INTERNATIONAL SYMPOSIUM ON THEORY OF MACHINES AND MECHANISMS BUCHAREST, ROMANIA, SEPTEMBER 1 - 4, 2005 SYROM 2005
DETERMINING THE MECHANICAL EFFICIENCY OF OTTO ENGINE’S MECHANISM Relly Victoria PETRESCU Florian Ion PETRESCU
Department of Descriptive Geometry and Engineering Graphics, University “POLITEHNICA” of Bucharest, 313 Splaiul Independentei, 77206 Bucharest, ROMANIA E-mail: victoriap@emoka.ro Department of Mechanisms and Robots, University “POLITEHNICA” of Bucharest, 313 Splaiul Independentei, 77206 Bucharest, ROMANIA E-mail: tiberiuionro@yahoo.com
ABSTRACT: The paper presents a few original elements about the dynamics and kinematics of piston mechanism, used like motor mechanism from OTTO engines. One presents an original method to determine the efficiency of the piston mechanism, used like motor mechanism. This method consists of eliminating the friction modulus. One determines the efficiency of the piston mechanism in two ways: 1. When the piston mechanism works like a motor; 2. When the piston mechanism works like a steam roller. Finally one determines the total motor efficiency, for the four cycle engine and for two cycle engine. With the relation of motor efficiency one optimizes the Otto mechanism, which is the principal mechanism from the internalcombustion engines. This is the way to diminish the acceleration of the piston and to maximize the efficiency of motor mechanism. One optimizes the constructive parameters: e, r, l, taking into account the rotation speed of drive shaft, n. Keywords: Efficiency, force, piston, crank, connecting-rod, motor, stroke, bore.
1. INTRODUCTION In this paper the authors present an original method to calculate the efficiency of the planar mechanisms with rods, concretely to the motor planar mechanism. The originality consists of eliminating the friction forces and friction coefficients. One determines, directly the mechanical efficiency of planar rods mechanism only. One determines the efficiency of piston mechanism (§ 2.) in two ways: 1. When the piston mechanism works like a motor; 2. When the piston mechanism works like a steam roller. Finally one determines the total motor efficiency, for the four cycle engine and for two cycle engine.
2. DETERMINING THE MECHANICAL MOTOR EFFICIENCY In figure 1 one can see the kinematical diagram of the mechanism with crank - connecting rod - piston [1,2]. The constructive parameters are: r, the radius of crank; l, the length of connecting-rod; e, the eccentricity between centre of crank rotation and axis of piston guide. The mechanism is positioned by the angle, ϕ, which is representing the rotation angle of crank. The connecting rod is positioned by one of the two angles, α or ψ (see picture 1). The variable length between the centre of crank rotation and the piston centre is yB. 2.1. The kinematics of Otto mechanism The kinematical relations (see fig. 1) are the following: ⎧r ⋅ cos ϕ + l ⋅ cosψ = −e ⎨ ⎩r ⋅ sin ϕ + l ⋅ sin ψ = y B
(1)
y B
3 2
0
l ψ yB
α
A ω
r ϕ
1
O
P
x 0
e
Fig. 1. The kinematical schema of Otto-mechanism From first relation of the positions system (1) one determines the value of ψ angle (see the relation 2): cosψ = −
e + r ⋅ cos ϕ l
(2)
From the second relation of system (1) one calculates directly the piston’s displacement, s=yB (see the relation 3): s = y B = r ⋅ sin ϕ + l ⋅ sinψ
(3)
One derivates the position’s system (1) and obtains the velocity’s system (4): ⎧− r ⋅ ϕ& ⋅ sin ϕ − l ⋅ψ& ⋅ sin ψ = 0 ⎨ ⎩r ⋅ ϕ& ⋅ cos ϕ + l ⋅ψ& ⋅ cosψ = y& B
(4)
From the first relation of system (4) one calculates the angular velocity, ψ& , (see the relation 5) and from the second relation of system (4) one determines the piston’s linear velocity, y& B , (see the relation 6):
ψ& = −
r ⋅ sin ϕ ⋅ ϕ& l ⋅ sinψ
y& B = r ⋅ ϕ& ⋅ cosϕ + l ⋅ψ& ⋅ cosψ
(5) (6)
One derivates the velocity’s system (4) and obtains the acceleration’s system (7): ⎧⎪− r ⋅ ϕ& 2 ⋅ cos ϕ − l ⋅ψ& 2 ⋅ cosψ − l ⋅ψ&& ⋅ sin ψ = 0 ⎨ ⎪⎩− r ⋅ ϕ& 2 ⋅ sin ϕ − l ⋅ψ& 2 ⋅ sin ψ + l ⋅ψ&& ⋅ cosψ = &y&B
(7)
From the first relation of system (7) one calculates the angular acceleration, ψ&& , (see the relation 8) and from the second relation of system (7) one determines the piston’s linear acceleration, &y&B , (relation 9):
ψ&& = −
r ⋅ ϕ& 2 ⋅ cos ϕ + l ⋅ ψ& 2 ⋅ cosψ l ⋅ sin ψ
(8)
(9)
&y& B = l ⋅ψ&& ⋅ cos ψ − r ⋅ ϕ& 2 ⋅ sin ϕ − l ⋅ψ& 2 ⋅ sin ψ
The angle α can be put in a function of ψ angle, see the expression (10):
α = ψ − 90
(10)
From the system (11) one can now determine the trigonometric functions of the α angle: ⎧cos α = sin ψ ⎨ ⎩sin α = − cosψ
(11)
With the expression (2) and the second relation of system (11), one determines sin α , see the relation (12):
sin α =
e + r ⋅ cos ϕ l
(12)
The piston’s velocity takes the form (13): r ⋅ ϕ& ⋅ sin ϕ ⋅ cosψ = sinψ r ⋅ ϕ& sin(ψ − ϕ ) sin(ψ − ϕ ) = ⋅ (cos ϕ ⋅ sinψ − sin ϕ ⋅ cosψ ) = r ⋅ ϕ& ⋅ = r ⋅ω ⋅ sinψ sinψ sinψ
vB = y& B = r ⋅ ϕ& ⋅ cos ϕ + l ⋅ψ& ⋅ cosψ = r ⋅ ϕ& ⋅ cos ϕ −
(13)
2.2. Determining the mechanical efficiency when the Otto mechanism works like a motor mechanism The Otto mechanism works like a motor mechanism in a single cycle (π angle), when the piston is moving from the near dead point to the distant dead point (when the piston is moving from an extreme position to another – see the picture 2); this is the single motor time. y near dead point
BI
3
0
y
0
l
3 l
l+r
2 ψI
αI
distant dead point
BII
2 AI r
e
αII
x
O P
l-r
1 ϕI 0
a - the crank is in prolonging with the connecting-rod
ω
P
1
ϕII
l e
ω
O 0
x r
ψII
AII b - the crank is overlapped on the connecting-rod
Fig. 2. The kinematical diagrams of Otto-mechanism in the extremely positions; a) when the crank is in prolonging of the connecting-rod, b) when the crank is overlapped on the connecting-rod
The efficiency of the pistons mechanism when the piston works like a motor mechanism can be determined, if one goes from the piston to the crank, with the determining of forces (see the figure 3), [2,3]. The consumed motor force (the input force), Fm, is divided in two components: 1) Fn - the normal force (in the long of the connecting-rod); 2) Fτ - the tangential force (perpendicular in B, on the connecting-rod); see the system (14); (in figure 3 ω is negative; the rotation sense is hourly). ⎧ Fn = Fm ⋅ cos α = Fm ⋅ sin ψ ⎨ ⎩ Fτ = Fm ⋅ sin α = − Fm ⋅ cosψ
(14)
Fn is the single force transmitted from B to A (because the rod hasn’t a joint with the frame).
y B Fn
α Fτ
ψ-ϕ
α l
ψ
yB Fm
ϕ
A
α ω
Fc ψ-ϕ r ϕ
O P
e
ψ-ϕ Fu Fn
x
0
Fig. 3. The forces of Otto-mechanism, when the piston works like a motor mechanism In A, the force Fn is divided in two components too: 1. Fu – the utile force; 2. Fc – a compression force. See the system (15): ⎧ Fu = Fn ⋅ sin(ψ − ϕ ) = Fm ⋅ sin ψ ⋅ sin(ψ − ϕ ) ⎨ ⎩ Fc = Fn ⋅ cos(ψ − ϕ ) = Fm ⋅ sin ψ ⋅ cos(ψ − ϕ )
(15)
The utile power, Pu, can be written in form (16): Pu = Fu ⋅ v A = Fu ⋅ r ⋅ ω = Fm ⋅ r ⋅ ω ⋅ sinψ ⋅ sin(ψ − ϕ )
(16)
The consumed power, Pc, can be written in form (17): Pc = Fm ⋅ vB = Fm ⋅ r ⋅ ω ⋅
sin(ψ − ϕ ) sinψ
(17)
The momentary mechanical efficiency, ηi, can be written with the relation (18):
ηi =
Pu F ⋅ r ⋅ ω ⋅ sin ψ ⋅ sin(ψ − ϕ ) (e + r ⋅ cos ϕ ) 2 = m = sin 2 ψ = cos 2 α = 1 − 1 Pc l2 Fm ⋅ r ⋅ ω ⋅ sin(ψ − ϕ ) ⋅ sin ψ
(18)
To calculate the mechanical efficiency, η, one can integrate the momentary efficiency, ηi, from near dead point to distant dead point, from ϕI to ϕII (figure 2 ): e ⎧ I ⎪⎪ϕ ≡ ϕ i = π − a cos( l + r ) ⎨ ⎪ϕ II ≡ ϕ = 2 ⋅ π − a cos( e ) f ⎪⎩ l−r
(19)
One determines approximately the efficiency with the relation (20), only if we can determine precisely the extreme angles, αM and αm:
η = 0.5 +
sin α M cos α M − sin α m cos α m 2 ⋅ (α M − α m )
(20)
2.3. Determining the mechanical efficiency when the Otto mechanism works like steam roller The Otto mechanism works like motor mechanism in a single cycle (a π angle), when the piston is moving from the near dead point to the distant dead point, and it works like steam roller in the rest of the energetically cycle. At the two cycle engines, the motor works like steam roller, in a single cycle, when the piston is moving from the distant dead point to the near dead point. At the four cycle engines, the motor works like steam roller, in three cycle; two times the piston is moving from the distant dead point to the near dead point, and in one cycle (one time) the piston is moving from the near dead point to the distant dead point. By a cycle (a π angle), one understands a time, a single time, precisely a semi kinematical-cycle; a kinematical cycle has a 2.π angle. In figure 4 one can see the forces in Otto mechanism when the mechanism works like a steam roller. Fn
Fu α
Fr
y
α B l Fn
α
ψ-ϕ
Fm ψ-ϕ
yB
ψ ψ-ϕ
α
Fτ ω e
A r ϕ
O P
ϕ
x 0
Fig. 4. The forces of Otto-mechanism, when the piston works like a steam roller The input force (the consumed motor force), Fm, perpendicular in A on the crank OA (r), is divided in two components: 1. Fn–the normal force, which is the active component, the only components transmitted from couple A to joint B; 2. Fτ–the tangential force, which can give a couple, and can rotate the connecting-rod, or bend it, [2,3]; see the system (21):
⎧ Fn = Fm ⋅ sin(ψ − ϕ ) ⎨ ⎩ Fτ = Fm ⋅ cos(ψ − ϕ )
(21)
In joint B, the transmitted force, Fn, is divided in two components too: 1. Fu – the useful force; 2. Fr – a force normal at the guide axis; see the system (22): ⎧ Fu = Fn ⋅ cos α = Fn ⋅ sin ψ = Fm ⋅ sin(ψ − ϕ ) ⋅ sin ψ ⎨ ⎩ Fr = Fn ⋅ sin α = − Fn ⋅ cosψ = − Fm ⋅ sin(ψ − ϕ ) ⋅ cosψ
(22)
The utile power can be written in form (23) and the consumed power can be written in form (24): Pu = Fu ⋅ vB = Fm ⋅ sin(ψ − ϕ ) ⋅ sinψ ⋅
rω sin(ψ − ϕ ) = Fm ⋅ r ⋅ ω ⋅ sin 2 (ψ − ϕ ) sinψ
Pc = Fm ⋅ v A = Fm ⋅ r ⋅ ω
(23) (24)
The momentary mechanical efficiency when the piston works like steam roller, can be calculated with the relation (25):
ηi = =
Pu Fm ⋅ r ⋅ ω ⋅ sin 2 (ψ − ϕ ) = = sin 2 (ψ − ϕ ) = Pc Fm ⋅ r ⋅ ω
[ l 2 − (e + r ⋅ cos ϕ ) 2 ⋅ cos ϕ + (e + r ⋅ cos ϕ ) ⋅ sin ϕ ] 2
(25)
l2
3. CONCLUSIONS The momentary mechanical efficiency when the piston works like steam roller (25), is different that the efficiency when the piston works like motor (18), (in the motor time). Generally the steam roller efficiency is lower that the motor efficiency. The steam roller efficiency is approximately 50% or a lower value and the motor efficiency can be 60-99%, in function of the constructive parameters, e, r, l. The motor efficiency increases when the ratio, λ=r/l, decreases. For a λ<0.33, the motor efficiency is high enough. The efficiency when the piston works like a steam roller may be calculated by integrating the momentary efficiency (25) in two ways: a) from ϕI to ϕII; b) from ϕII to ϕI. Generally the results are not the same. One must calculate three types of efficiency for the four cycle engines, and one should calculate two ways of efficiency for the two cycle engines. The final efficiency for the four cycle engines can be 60%, and for the two cycle engines can be 73%, with e=10 [mm], r=20 [mm], l=90 [mm] for example (see figure 1). The two cycle engines can give us a 13% more efficiency (this fact may be important). The Otto mechanism may be improved for giving a better efficiency and a minimum value for the maximum acceleration [3]. Constructive, one must adopt a lower stroke and a greater bore. The radius of crank, r, must be shorter. The piston should take the aspect of a pot (a frying pan).
REFERENCES 1. Pelecudi, Chr., ş.a., Mecanisme, E.D.P., Bucureşti, 1985; 2. Petrescu, V., Petrescu, I., Randamentul cuplei superioare de la angrenajele cu roţi dinţate cu axe fixe, In: The Proceedings of 8th National Symposium PRASIC, Braşov, vol. I, pp. 333-338, 2002. 3. Petrescu, F.I., Petrescu, R.V., Câteva elemente privind îmbunătăţirea designului mecanismului motor, In: The Proceedings of 8th National Symposium on GTD, Braşov, vol. I, pp. 353-358, 2003.