TECHNICAL UNIVERSITY OF CLUJ-NAPOCA ACTA TECHNICA NAPOCENSIS
International Conference on Engineering Graphics and Design 12-13 June 2009
MF1 DESIGN Florian Ion PETRESCU, Relly Victoria PETRESCU Abstract: The paper presents a new and original internal-combustion engine. It is presenting a method in determining the kinematics and the efficiency of a new mechanism, MF1, proposed (designed) to work and be tested like an internal-combustion engine. One determines the mechanical momentary efficiency, when the mechanism works like a steam roller and when the mechanism works like a motor. The determined efficiency is different in the two described situations. One presents an original way to determine the dynamic efficiency too. The dynamic momentary efficiency is the same in the two situations: when the mechanism works like a steam roller and when it works like a motor. One determines the efficiency without friction, but one can anytime add the effect of friction modulus. One presents the dynamic kinematics of this mechanism too: the dynamic velocity and the dynamic acceleration. When the constructive parameters are normal, the dynamic velocities take the same values like the classical speeds and the dynamic accelerations take the same values like the classical accelerations. Key words: Efficiency, force, piston, crank, connecting-rod, motor, stroke, bore, dynamic-velocity, dynamic-acceleration, dynamic-efficiency.
1. INTRODUCTION The paper presents shortly a new and original internal-combustion engine, MF1. The originality consists in the way of determining the mechanical and dynamic efficiency and in the way of determining the dynamic velocities and accelerations. In this paper one determines the efficiency of piston mechanism in two ways: 1.When the piston mechanism works like a motor; 2.When the piston mechanism works like a steam roller. 2. PRESENTING THE KINEMATICS OF MF1 In the picture number 1, one can see the kinematics outline of the new presented motor mechanism (Motor, Florio 1), [2]. The first modification of this model, having in view the classical model (Otto engine mechanism), is the use of two connecting-rod, (2 and 3) and the use of B couple, a dual couple: of rotation and translation.
This motor mechanism is a new mechanism and his functionality will be different from the classical mechanism’s functionality. The great advantage of this mechanism is that it can be regulated to have a bigger zone with constant acceleration at the piston (the element number five). The efficiency of this mechanism is the same like the Otto mechanism. The structural group 2-4 (a dyad) can improve the motor functionality without damage of power. The kinematics relations are the following (1-11): a 2 = l 02 + l12 − 2 ⋅ l 0 ⋅ l1 ⋅ sin ϕ 1 l ⋅ cos ϕ 1 cos ϕ 2 = − 1 a e − l1 ⋅ cos ϕ 1 − l 2 ⋅ cos ϕ 2 cos ϕ 3 = l3 y D = l1 ⋅ sin ϕ 1 + l 2 ⋅ sin ϕ 2 + l 3 ⋅ sin ϕ 3 l ⋅ cos(ϕ 1 − ϕ 2 ) ω2 = − 1 ⋅ ω1 a l ⋅ l ⋅ ω ⋅ cos ϕ 1 a& = − 0 1 1 a
(1) (2) (3) (4) (5) (6)
ω3 =
l1 ⋅ ω1 ⋅ [l 0 ⋅ cos ϕ1 ⋅ sin(ϕ 3 − ϕ 2 ) + a ⋅ l3
+ b ⋅ cos(ϕ1 − ϕ 2 ) ⋅ cos(ϕ 3 − ϕ 2 )] y& D = ω1 ⋅ l1 ⋅ cos ϕ1 +
(8)
+ ω 2 ⋅ l 2 ⋅ cos ϕ 2 + ω 3 ⋅ l 3 ⋅ cos ϕ 3 y
MOTOR FLORIO1-MF1 © 2003 Florian PETRESCU The Copyright-Law Of March, 01, 1989 U.S. Copyright Office Library of Congress Washington, DC 20559-6000 202-707-3000
(7)
5 D
3
0
a ⋅ Fm ⋅ cos(ϕ 2 − ϕ1 ) ⋅ sin(ϕ 2 − ϕ 3 ) = b = Fm ⋅ [sin(ϕ 2 − ϕ1 ) ⋅ cos(ϕ 2 − ϕ 3 ) +
e
b
+
yD
B 4 0
1
a l0 O
l2 l1
A
(13) (14)
= Fm ⋅ sin(ϕ 2 − ϕ1 ) ⋅ cos(ϕ 2 − ϕ 3 ) +
C 2
(12)
+ Fτ C ⋅ sin(ϕ 2 − ϕ 3 ) =
l3 φ3
⎧Fn = Fm ⋅ sin(ϕ 2 − ϕ1 ) ⎨ ⎩Fτ A = Fm ⋅ cos(ϕ 2 − ϕ1 ) a a Fτ C = ⋅ Fτ A = ⋅ Fm ⋅ cos(ϕ 2 − ϕ 1 ) b b I ⎧⎪ Fn = Fn ⋅ cos(ϕ 2 − ϕ 3 ) ⎨ I ⎪⎩ Fτ C = Fτ C ⋅ sin(ϕ 2 − ϕ 3 ) FT = FnI + Fτ IC = Fn ⋅ cos(ϕ 2 − ϕ 3 ) +
ϕ2
φ1
+ x
0
Fig. 1. The MF1 kinematics outline
l1 ⋅ ω1 ⋅ (ω1 − ω 2 ) ⋅ sin(ϕ 1 − ϕ 2 ) − ω 2 ⋅ a& (9) a l ⋅ω ε 3 = 1 1 ⋅ [ −l 0 ⋅ ω1 ⋅ sin ϕ1 ⋅ sin(ϕ 3 − ϕ 2 ) + a ⋅ l3 + l 0 ⋅ (ω 3 − ω 2 ) ⋅ cos ϕ1 ⋅ cos(ϕ 3 − ϕ 2 ) − − b ⋅ (ω1 − ω 2 ) ⋅ sin(ϕ1 − ϕ 2 ) ⋅ cos(ϕ 3 − ϕ 2 ) − (10) − b ⋅ (ω 3 − ω 2 ) ⋅ cos(ϕ1 − ϕ 2 ) ⋅ sin(ϕ 3 − ϕ 2 )] − a& − ⋅ ω3 a &y&D = −ω12 ⋅ l1 ⋅ sin ϕ1 − ω 22 ⋅ l 2 ⋅ sin ϕ 2 +
ε2 =
+ ε 2 ⋅ l 2 ⋅ cos ϕ 2 − ω 32 ⋅ l 3 ⋅ sin ϕ 3 +
a ⋅ cos(ϕ 2 − ϕ1 ) ⋅ sin(ϕ 2 − ϕ 3 )] b ⎧ FU = FT ⋅ sin ϕ 3 ⎨ ⎩ FR = FT ⋅ cos ϕ 3 ⋅ cos(ϕ 2 − ϕ 3 ) + +
(17)
a ⋅ cos(ϕ 2 − ϕ1 ) ⋅ sin(ϕ 2 − ϕ 3 )] b FU
FT
ϕ2 ϕ2-ϕ3
Fn C
FR
φ3 0
b
B Fn ϕ2-ϕ1 a
Fm MOTOR FLORIO1-MF1 © 2003 Florian PETRESCU The Copyright-Law Of March, 01, 1989 U.S. Copyright Office Library of Congress Washington, DC 20559-6000 202-707-3000
φ3
D
FτC
+ ε 3 ⋅ l 3 ⋅ cos ϕ 3
One can determine the momentary mechanical efficiency, when the mechanism works like a steam roller, if one determines the distribution of forces, from the crank to the piston (figure 2); relations (12-19) [2]:
(16)
FU = Fm ⋅ sin ϕ 3 ⋅ [sin(ϕ 2 − ϕ1 ) ⋅
(11)
3. DETERMINING THE MOMENTARY MECHANICAL EFFICIENCY WHEN THE MECHANISM WORKS LIKE A STEAM ROLLER
(15)
FτA
ϕ2 A φ1
O
φ1
x
Fig. 2. The MF1 distribution of forces, when the mechanism works like a steam roller l2 ⋅ cos ϕ 2 ⋅ cos(ϕ1 − ϕ 2 ) + a l0 ⋅ cos ϕ1 ⋅ sin(ϕ3 − ϕ 2 ) + b ⋅ cos(ϕ1 − ϕ 2 ) ⋅ cos(ϕ3 − ϕ 2 ) (18) a ⋅ cos ϕ3 ] vU = vm ⋅ [cos ϕ1 −
η iC =
Fu = Fu1 + Fu 2 = Fm ⋅ sin ϕ 3 ⋅
FU ⋅ vU = sin ϕ 3 ⋅ [sin(ϕ 2 − ϕ1 ) ⋅ cos(ϕ 2 − ϕ 3 ) + Fm ⋅ v m
a ⋅ cos(ϕ 2 − ϕ1 ) ⋅ sin(ϕ 2 − ϕ 3 )] ⋅ b (19) l ⋅ cos ϕ 2 ⋅ cos(ϕ1 − ϕ 2 ) ⋅ [cos ϕ1 − 2 + a l 0 ⋅ cos ϕ1 ⋅ sin(ϕ 3 − ϕ 2 ) + b ⋅ cos(ϕ1 − ϕ 2 ) ⋅ cos(ϕ 3 − ϕ 2 ) a ⋅ cos ϕ 3 ]
⋅ [cos(ϕ 2 − ϕ 3 ) ⋅ sin(ϕ1 − ϕ 2 ) −
+
4. DETERMINING THE MOMENTARY MECHANICAL EFFICIENCY WHEN THE MECHANISM WORKS LIKE A MOTOR One can determine the momentary mechanical efficiency, when the mechanism works like a motor, if one determines the distribution of forces, from the piston to the crank (figure 3); relations (20-25) [2]: MOTOR FLORIO1-MF1 © 2003 Florian PETRESCU The Copyright-Law Of March, 01, 1989 U.S. Copyright Office Library of Congress Washington, DC 20559-6000 202-707-3000
ϕ3
D ϕ2-ϕ3
FτC
ϕ2
FN φ3
FR
Fm
C
0
b
FN B A
ϕ3
a
Fn
ϕ2 φ1
Fu2
FτA ϕ1-ϕ2
Fu1
Fn
φ1 O
x
Fig. 3. The MF1 distribution of forces, when the mechanism works like a motor
⎧ FN = Fm ⋅ sin ϕ 3 (20) ⎨ ⎩ FR = Fm ⋅ cos ϕ 3 ⎧Fn = FN ⋅ cos(ϕ 2 − ϕ 3 ) (21) ⎨ ⎩Fτ C = FN ⋅ sin(ϕ 2 − ϕ 3 ) b b Fτ A = ⋅ Fτ C = ⋅ Fm ⋅ sin ϕ 3 ⋅ sin(ϕ 2 − ϕ 3 ) (22) a a ⎧ Fu1 = Fn ⋅ sin(ϕ1 − ϕ 2 ) (23) ⎨ F F ϕ ϕ = − ⋅ cos( − ) u 2 τ 1 2 A ⎩
−
(24)
b ⋅ sin(ϕ 2 − ϕ 3 ) ⋅ cos(ϕ1 − ϕ 2 )] a
⎧ N = − sin ϕ 3 ⋅ [cos(ϕ 2 − ϕ 3 ) ⋅ sin(ϕ1 − ϕ 2 ) − ⎪ b ⎪− ⋅ sin(ϕ 2 − ϕ 3 ) ⋅ cos(ϕ1 − ϕ 2 )] ⎪ a ⎪n = cos ϕ + cos ϕ ⋅ 1 3 (25) ⎪ ⎪ l 0 cos ϕ1 sin(ϕ 3 − ϕ 2 ) + b cos(ϕ1 − ϕ 2 ) cos(ϕ 3 − ϕ 2 ) ⎨ a ⎪ ⎪ l 2 cos ϕ 2 cos(ϕ1 − ϕ 2 ) ⎪− a ⎪ ⎪ N ⎪μ iM = n ⎩
5. DETERMINING THE MOMENTARY DYNAMIC EFFICIENCY The dynamic efficiency of the mechanism is the same, anytime (when the mechanism works like a steam roller and when it’s working like a motor). It can be determined approximately with the relation (26): D ⎧μ iD = μ iM = η iCD = sin 2 ϕ 3 ⋅ sin 2 τ ⎪ (26) ⎨ ⎪⎩with : τ = 2 ⋅ ϕ 2 − ϕ 1 − ϕ 3 One can determine the exactly momentary dynamic efficiency of the mechanism, if one takes in calculation the dynamic velocities (in this case the speeds distribution is the same like the forces distribution), see the relations (2729): −
Fu = sin ϕ 3 ⋅ [sin(ϕ 2 − ϕ1 ) ⋅ cos(ϕ 2 − ϕ 3 ) Fm
b + ⋅ sin(ϕ 2 − ϕ 3 ) ⋅ cos(ϕ 2 − ϕ1 )] a v − u = sin ϕ 3 ⋅ [sin(ϕ 2 − ϕ1 ) ⋅ cos(ϕ 2 − ϕ 3 ) vm a + ⋅ sin(ϕ 2 − ϕ 3 ) ⋅ cos(ϕ 2 − ϕ1 )] b μ iD = sin 2 ϕ 3 ⋅ {sin 2 (ϕ 2 − ϕ1 ) ⋅ cos 2 (ϕ 2 − ϕ 3 ) + + sin 2 (ϕ 2 − ϕ 3 ) ⋅ cos 2 (ϕ 2 − ϕ1 ) + +
a2 + b2 ⋅ sin[ 2 ⋅ (ϕ 2 − ϕ1 )] ⋅ sin[ 2 ⋅ (ϕ 2 − ϕ 3 )]} 4⋅a ⋅b
(27)
(28)
(29)
6. THE DYNAMIC KINEMATICS OF THE MECHANISM One can determine now the dynamic velocity (30) and the dynamic acceleration of the piston (31): v DDin = l1 ⋅ ω1 ⋅ sin ϕ 3 ⋅ [sin(ϕ 2 − ϕ1 ) ⋅ cos(ϕ 2 − ϕ 3 ) a + ⋅ sin(ϕ 2 − ϕ 3 ) ⋅ cos(ϕ 2 − ϕ1 )] b a
Din D
(30)
100
-150000
200
300
400
aD(D) aD(Cin)
-200000
(31)
a ⋅ sin(ϕ 2 − ϕ1 ) ⋅ sin(ϕ 2 − ϕ 3 ) ⋅ (ω 2 − ω1 ) + b
a + ⋅ cos(ϕ 2 − ϕ1 ) ⋅ cos(ϕ 2 − ϕ 3 ) ⋅ (ω 2 − ω 3 ) + b a& + ⋅ cos(ϕ 2 − ϕ1 ) ⋅ sin(ϕ 2 − ϕ 3 ) + b a ⋅ a& + 2 ⋅ cos(ϕ 2 − ϕ1 ) ⋅ sin(ϕ 2 − ϕ 3 )]} ⋅ l1 ⋅ ω1 b
Fig. 5. The kinematical and dynamic accelerations
8. CONCLUSION Some mechanisms have the same parameters for the classical and for the dynamic kinematics (gears, cams with plate followers). At the presented mechanism, the dynamic-kinematics is different from the classical-kinematics, but, if the constructive parameters are normal, the dynamic velocities and accelerations practically take the same values like the classical speeds and accelerations.
7. DISCUTION n=5000[rot/min];l1=0.08;l2=0.3;l3=0.5;l0=0.15;e=0
2
V(Din)/(l1.w) V(Cin)/(l1.w)
1
0 -50000 -100000
a ⋅ cos(ϕ 2 − ϕ1 ) ⋅ sin(ϕ 2 − ϕ 3 )] + sin ϕ 3 ⋅ [cos(ϕ 2 − ϕ1 ) ⋅ b ⋅ cos(ϕ 2 − ϕ 3 ) ⋅ (ω 2 − ω1 ) − sin(ϕ 2 − ϕ1 ) ⋅ sin(ϕ 2 − ϕ 3 ) ⋅
1.5
n=5000[rot/min];l1=0.08;l2=0.3;l3=0.5;l 0=0.15;e=0
50000
0
= {ω 3 ⋅ cos ϕ 3 ⋅ [sin(ϕ 2 − ϕ1 ) ⋅ cos(ϕ 2 − ϕ 3 ) +
⋅ (ω 2 − ω 3 ) −
of the piston (30-31), are not the same like the classical kinematics values (8-11), see the pictures number (4 and 5):
0.5 0 -0.5 0
100
200
300
400
9. REFERENCES
-1 -1.5 -2
Fig. 4. The kinematical and dynamic velocities
When the values of the constructive parameters are different from the normal, the dynamic speeds and the dynamic acceleration
[1] Pelecudi, Chr., s.a. Mecanisme, E.D.P., Bucuresti, 1985. [2] Petrescu, F.I., Petrescu, R.V., An original internal combustion engine, Proceedings of 9th International Symposium SYROM, Vol. I, p. 135-140, Bucharest, 2005.
DESIGNUL MOTOARELOR MF1 Rezumat: Lucrarea prezintă un nou motor cu ardere internă de tip Otto. Se prezintă cinematica şi randamentul noului motor când mecanismul principal lucrează în regim de compresor şi de motor. Se determină deasemenea randamentul dynamic, viteza şi acceleraţia dinamică.
Authors: Petrescu Florian Ion, PhD. Eng. Assistant Professor at Polytechnic University of Bucharest, TMR Department (Theory of Mechanisms and Robots Department), petrescuflorian@yahoo.com, 0214029632; Petrescu Relly Victoria, PhD. Eng., Lecturer at Polytechnic University of Bucharest, GDGI Department (Department of Descriptive Geometry and Engineering Graphics), petrescurelly@yahoo.com, 0214029136.