The 2nd International Conference
Computational Mechanics
and
Virtual Engineering
COMEC 2007
11 – 23 OCTOBER 2007, Brasov, Romania
THE COUPLING SYSTEMS OF THE POWER SOURCES,
USED AT HYBRID AUTOMOBILE
Nicolae PANDREA1 and Dinel POPA2
1
University of Pitesti, ROMANIA, nicolae_pandrea37@yahoo.com
2
University of Pitesti, ROMANIA, dinel_popa@yahoo.com
Abstract: In this paper is presented one solution for coupling the power sources. Are made the movement differential equations for
the planetary mechanisms with two degrees of mobility and by using them the movement is studied. For the permanent movement is
studied the mechanisms frequently used and then is studied the movement stability. In the end of the paper are presented three
numerical applications.
Keywords: power source, hybrid drive, permanent movement
1. INTRODUCTION
The hybrid automobiles haulage is made with a heat engine (MT) and an electric engine (ME) that is powered to a
battery of a high tension accumulation.
The battery is charged from an electric generator that is powered by the heat engine. The power sources can be coupled
through a planetary mechanism with two degrees of mobility. The general draw of such a system is presented in fig. 1.
1
z'4
z"4
O
4
B
2
A
3
O1
O3
O2
z3
z2
MT
GE
BA
ME
Fig. 1. The planetary mechanism.
2. MOVEMENT DIFFERENTIAL EQUATION OF THE COUPLING SYSTEM
2.1. The notations
One considered the notations:
- J i , i  1, 2, 3, 4 , axial moments of inertia of the elements marked with 1, 2, 3, 4;
- m4 , weight of the satellites 4;
- R4 , the length of the port-satellite 1;
- z2 , z3 , z4' , z4" the number of teeth of the tooth wheels;
- i1 , i the equations defined by the relations
537
z2
z z"
; i  2 4' ;
'
z4
z3 z4
- A, B, C, parameters inertial parameters defined by the relations
A  J1  m4 R2  1  i12 J 4  1  i J3 ; B  1  i iJ3  J 4 1  i1 i1 ; C  J 2  J3i 2  J 4i12 ;
i1 


(1)
(2)
- 1 ,  2 , 3 the rotation angles of the elements 1, 2, 3;
- 1, 2 , 3 , 4 the absolute angular velocities of the elements 1, 2, 3, 4;
- M1 , M 2 , M 3 moments that act on the elements 1, 2, 3.
2.2. The kinetic energy and the generalized forces
Using the Willis method one read the relations
4  1
z   1
z"
  2' ; 3
 4.
2  1
z3
z4 4  1
(3)
from which, with the notations (1) he or she deduced the equalities:
3  11  i   2i ; 4  11  i1   2i1 .
The kinetic energy of the system:
1
EC  ( J112  m4 R 212  J 222  J 332  J 424 ) .
2
with the notations (4) and (2) has the equation:
1
EC  A 12  2 B 1 2  C 22 . (6)
2
The mechanic power at a certain time is given by:
P  M11  M 22  M 33 ,
or on the basis of the other relation (4)
P  M1  M 3 (1  i)1  (M 2  M 3i)2
and from here are duced the generalized equations:
Q1  M1  M 3 (1  i) ; Q2  M 2  M 3i .

(4)
(5)

(7)
(8)
(9)
2.3. Lagrange equations
Knowing the fact that
1  1 ; 2   2 ,
from Lagrange equations are obtained the differential equations
 1  B
 2  M1  M 3 (1  i);  B
 1  C
 2  M 2  M 3i ,
A
or
M  M 3 1  i C  M 2  M 3i B ;
1  1

AC  B 2
M  M 3 1  i B  M 2  M 3i A .
2  1

AC  B 2
(10)
(11)
(12)
3. THE MOVEMENTS STUDY
3.1. Permanent movement
The permanent movement is deduced from the conditions
1  0 ; 2  0 ,
that leads to these equations
M1  M 3 (1  i)C  (M 2  M 3i)B  0 ; M1  M 3 (1  i)B  (M 2  M 3i) A  0
and since AC  B  0 , we obtained the equations
M1  M 3 (1  i)  0 ; M 2  M 3i  0 .
(13)
(14)
2
(15)
The sources of power are present by the characteristics movement-angularly speed through the relations M i  M i (i )
538
and goes to the values: 1* , *2 , *3 of the angularly velocities that are obtained the permanent condition that are
deduced from the equation system:
M1(1* )  (1  i)M3 (*3 )  0 ;
M 2 (*2 )  i M 3 (*3 )  0 ;
*3
If these values
relations:
 i*2 .
M 2 (*2 ) ,
(16)
 (1  i)1*
M 3 (*3 ) are related to the value of the moment M1(1* ) of the heat engine, one obtained the
M 2 (*2 )
i
M (* )
1
; 3 *3  
,

*
i

1
i

1
M1 (1 )
M1 (1 )
with the graphic representations captured in figure 2 and figure 3.
(17)
M 2 (*2 )
M 1 (1* )
4
3
3
2
2
1
-4
-3
-2
1
-1
-4
1
2
3
4
5
-3
-2
-1
i
1
-1
-1
-2
-2
-3
-3
-4
-4
Fig. 2. Graphic
M 2 (*2 )
M 1 (1* )
M 2 (*2 )
i
.

*
i 1
M1 (1 )
Fig. 3. Graphic
2
3
4
5
i
M 3 (*3 )
1
.

*
i 1
M1 (1 )
The displacement moment M 2 (*2 ) , because the resistant moment is negative, it follows that the transmission equation
i will be from the domain (,0)  (1, ) .
| M 2 (2 ) | | M 3 (*3 ) |
are sub-unitary, if this 1  i  2 is over-unitary and this 2  i the first
,
M1 (1* ) M1 (1* )
value is over-unitary and the second one is sub-unitary.
In conclusion, for the solution where the values of the resistant moment are low then we’ll chose i  0 and for the case
when the resistant moment at the automobiles wheels are high will be obtained the solution i  2 .
If i  0 the values
3.2. The movements stability
The solution 1* , *2 , *3 is obtained from the (16) system and it is stabile even if i  i 0  *i , i  1, 2, 3 are going
through zero.
In this way will be studied the stability after the first approximation.
Are obtained the linear approximations:
M1(1*  1 )  M1 (1* )  1M1 p (1* ) ,
M 2 (*2   2 )  M 2 (*2 )  2 M 2 p (*2 )
(18)
M 3 (*3  3 )  M 3 (*3 )  3M 3 p (*3 )
where by M ip were noted the functions derivates M i (*i ) .
By replacing the equations (12) and taking into account the notations
1
i (1  i )
1

M1 p (1* )  (1  i ) 2 M 3 p (*3 ) ,  
M 3 p (*3 ) ,  
M 2 p (*2 )  i 2 M 3 p (*3 )
2
2
2
AC  B
AC  B
AC  B
one obtained the linearly system of differential equations
  (C  B)  (C  B) , 
  (B  A)  (B  A)

1
1
2
2
1
2
The characteristic equation reads



539

(19)
(20)
C   B  r
C  B
0
B   A
 B  A  r
(21)
or
r 2  Dr  E  0
where
D  (C  2B  A) , E  (C  B)(B  A)  (B  A)(C  B)
By making the calculations, from equations (23), (19) are obtained the results
1
D
CM1 p (1* )  AM 2 p (*2 )  i 2 A  2i (1  i ) B  (1  i ) 2 C M 3 p
2
AC  B
1
E
M1 p (1* ) M 2 p (*2 )  (1  i ) 2 M 2 p (*2 ) M 2 p (*2 )  i 2 M1 p M 3 p
2
AC  B
For the evaluation of the signs of the of the parameters D, E are taking into account the inequalities:


 


(22)
(23)
(24)
A  0; C  0; AC  B 2  0;
(25)
i 2 A  2i(1  i) B  (1  i)2 C  0
that are deduced from the element calculations from the relation (2)
Therefore, if M ip (*i ), i  1, 2, 3 are negative, then D  0 ; E  0 , the equation (22) has real negative solutions or
complex solutions with the real part negative and thus 1  0 ,  2  0 and the movement is stabile.
4. APPLICATIONS
4.1. Application 1
Case when de heat engine has the characteristic M1  M 01  2b11  c11* , M 01  0; b1  0; c1  0, the automobile is
*
*
moving with a constant resistant moment M 2  M 02
; M 02
 0 and the generator has the characteristic M 3  b33 ,
b3  0 .
In this case by eliminating the parameter M 3 (*3 ) from the system (16) is obtained the equation
i 1
i
with the graphic representation from figure 4.
*
M 01  2b1  c11*  M 02
(26)
M 01 i-1
i
M1
*1
1
Fig. 4. Graphic M1 (1 )
The angularly velocity 1* is

* i 1
b1  b12  c1 M 01  M 02

i 

1* 
c1
From the second equation (18) and from the relation (39) is obtained
*
M 02
ib3
and from the equation (16) it results
*3  
(27)
(28)
540
(i  1)1*  *3
.
i
For the stability study is first calculated the derivate:
M1 p (1* )  2(b1  c11* )
(30)
and by taking into account that the relation (27) is negative and then
M 2 p (*2 )  0; M 3 p (*3 )  b3  0
(31)
*2 
(29)
On the basis of these relations are deduced from (24) the inequalities
D  0; E  0
and as a result the solution is stabile.
The mechanical power that is produced by the heat engine is
* * i 1
P1  M11*  M 02
1
.
i
The mechanical powers that are consumed are
- the mechanical power from the axel-tree
* *
P2  M 02
2 (34)
the power consumed at the generator
M 
P 
3
* 2
02
i 2b3
.
(32)
(33)
(35)
The balance of the powers leads us to:
(36)
P1  P2  P3
In a numerical application, considering that the power P2 from the wheel is equal with 60kW and the angularly velocity
*2  100rad/s is obtained the resistant moment M 02  600 Nm.
Further considering the transmission equation 1*  250 rad/s and angularly velocity of the engine is 1*  250 rad/s Nm
*
results M 1  M 02
i 1
 360 the and the mechanical power is P1  90 kW.
i
The angularly velocity *3 is given by the relation *3  (1  i)1*  i*2  125rad/s .
*
M 02
 240 Nm and thus the power that is
i
M
consumed by the generator is P3  M 33  30 kW, and the equivalent constant of the generator is b3  *3  1,92 .
3
If we don’t like this constant value can be done an amplification of n times of the angularly velocity, mentioning the
~
b
240
same power and then the generator's relation is b3  32 and the moment is
Nm.
n
n
The moment from the generator is deduced from the relation (19) M 3 
4.2 . Application 2
When the heat engine has the characteristics: M1  M 01  2b11  c112 , M 01  0, b1  0, c1  0 , the automobile is moving
*
with a resisting moment M 02
 0 , the generator is functioning with the characteristics M 3  b33 and the heat engine
~
with M 2  M 02  b22 that helps for automobiles displacement.
From the equations (16) is deduced the equality
i 1
M1  
M2
(37)
i
*
and since M 2  M 02  M 02
 b22 it results that
i 1
*
( M 02  M 02
 b22 ) .
i
The second relation (16) is now deduced:
M 01  2b11  c112  
(38)
*
M 02  M 02
 b22  ib3  (i  1)1  i2   0
(39)
from where we obtain
541
*
M 02  M 02
 i(i  1)b31
b2  i 2b3
and the equation (38) is
*

(i  1)2 b2b3 
i(i  1)b3 (M 02  M 02
)
c112  2b1 


M

0.

1
01
2
2
b2  i b1 
b2  i b1

2 
(40)
(41)
From the relation (41) one obtained the angularly velocity 1* , from (40) the angularly velocity *2 , and *3 is obtained
from the same equation
(42)
*3  (i  1)1*  i*2
4.3. Application 3
The case when the heat engine has the characteristic: M1  M 01  2b11  c12 , M 01  0 , b1  0 , c1  0 , the
*
automobile is moving with a resistant moment M 02
 0 when the generator is stopped. The electric engine has the
~
characteristic M  M 02  b22 , that helps to the automobiles displacement.
In this case 3  0 the system has only one degree of mobility
i 1
1
i
and the total resistant moment is
*
M 2  M 02  M 02
 b22 ,
or
i 1
*
M 2  M 02  M 02

b21 . (45)
i
The generalized force in this case is
2 
(43)
(44)
i 1
M2,
i
the angularly velocity 1* is obtained from the equation
Q  M1 
(46)
i 1
M2  0 ,
i
or from the equation
M1 
M 0  2b11  c11* 
(47)
i 1
i 1

*
b21   0 ,
 M 02  M 02 
i 
i

(48)
and it follows that
2
2
2

i 1
 i  1  b2
 i  1  b2 

* 
b1  
 b1  
M 02  M 02
  c1  M 01 



i  2
i
c
i









*
1 
.
c1


(49)
REFERENCES
[1] Pandrea, N., Popa D.,: Mecanisme. Teorie şi aplicaţii CAD, Ed. Tehnică, Bucureşti, 2000.
[2] Popa, D., Chiroiu, V., L. Munteanu, J. Onisoru, A. Iarovici: On the modeling of hybrid systems, Proc. of the
Romanian Academy, Series A: Mathematics, Physics, Technical Sciences, Information Science, vol. 8, nr. 1, 2007.
[3] Teodorscu, P. P.: Sisteme mecanice. Modele clasice, Ed.Tehnică, Bucureşti, vol. 1-3, 1984-1988-1997.
[4] Dudiţă, FL: Mecanisme, Ed. Universitatea din Braşov, 1977.
542